Problem Set 5 Solutions

Transcription

1 Problem Set 5 Solutions Dorian Abbot APM47 0/8/04. (a) We consider π θ π. The pendulum pointin down corresponds to θ=0 and the pendulum pointin up corresponds to θ π. Define ν θ. The system can be rewritten: θ ν () ν sin θ ω cos θ R Notice that this system is reversible. The fied points of this system are at ν=0 and θ 0 π arcos. Calculate the Jacobian: Rω J ω cos θ sin θ The trace and determinant of the Jacobian are: Evaluate at the fied points: i. Rω < 0 cos θ 0 R τ 0 (3) () R cos θ ω sin θ cos θ (4) 0 0 ω π 0 ω arcos Rω 0 ω Rω (5) (6) Rω Rω (7) The f.p. at (0,0) is a saddle. The f.p. at ( π,0) is a saddle. The f.p. at ( arcos,0) Rω are nonlinear centers. ii. = Rω The f.p. at ( π,0) is a saddle. There is another f.p. at (0,0) which is a center (from plottin). iii. > Rω The f.p. at (0,0) is a center. The f.p. at ( π,0) is a saddle. The f.p. at ( arcos,0) Rω don t eist. (b) If we include friction our equations can be written: θ ν (8) µν () ν sin θ ω cos θ R

2 The system is no loner reversible, so we don t epect to et centers. We et the same fied points as before. The Jacobian is: J ω cos θ sin θ The trace and determinant of the Jacobian are: Now let s consider our cases aain: (a) (b) (c) 0 cos θ R µ τ µ (0) R cos θ ω sin θ cos θ () < Rω The f.p. at (0,0) is a saddle. The f.p. at ( π,0) is a saddle. The f.p. at ( arcos,0) Rω are either stable nodes or stable spirals. Whether τ 4 µ ω is reater R ω 4 than or less than zero determines whether these points will be nodes or spirals. Rω = At this point a supercritical pitchfork bifurcation occurs. > Rω The f.p. at ( π,0) is a saddle. The f.p. at (0,0) is either a stable spiral or a stable node. Whether τ 4 µ R ω is reater than or less than zero determines whether these points will be nodes or spirals. The f.p. at ( arcos,0) don t eist. Rω Here is a bifurcation diaram:

3 4 Bifurcation Diaram 3 saddle stable node or spiral θ 0 saddle ω/(/r) / 3

4 . First notice that the paper cited was written by Harvard s very own Professor Howard Stone! (a) The system is invariant under the transformation t t φ φ, so it is reversible. β (b) ẋ 0 at φ 0 π and =0, while φ 0 at 8 cos φ. Now consider the intersection of the nullclines. At φ π we can never et a f.p. with positive β and. At φ 0 there will be a f.p. at =8 β for β! 8. At =0 there will be fied points at φ " arccos β for 0 β 8!. At = there will β be fied points at φ # arccos for 0 β 8. So for there are three fied points. The Jacobian of the system is:! β! 8 J %$ The trace and determinant of the Jacobian are: τ At (φ,)=(0,8 β ), we have: So for! β! 8 4 sin φ 4 cos φ 6 cos φ sin φ 8 & 7 sin φ () sin φ ' At φ ( ) arccos 8 β *, we have: 64 cos φ (3) τ 0 (4) β β 8 (5) (when this f.p. eists) <0 and it is a saddle. τ 7 sin φ (6) 3 64 sin φ (7) sin φ So τ 4 sin φ So the f.p. at positive φ is an unstable node and the f.p. at neative φ is a stable node. At =0, ẋ=0 and for our β rane φ, 0, so a closed orbit winds around the cylinder. For near one, the stable node at φ -. ) arccos 8 β is approached at lon times. So there must be a homoclinic orbit that intersects with the saddle point and winds around the cylinder to separate these two domains. By the Poincare-Bendion theorem there must be at least one closed orbit between =0 and the homoclinic orbit. We epect to have non-isolated closed orbits in reversible nonlinear systems, so we epect to have an entire band of closed orbits in this reion. Here is a phase portrait: 4

5 = (sqrt()/4) ( ) sin(phi) phi = (/) (beta cos(phi)/sqrt() cos(phi)/(8 sqrt())) beta =.05/sqrt() 3 phi

6 (c) Define β α. So β from above corresponds to α 0. The position of the saddle is ( 8α,0), so the saddle moves toward the line =0 as α 0, shrinkin the ), the homoclinic orbit becomes the circle in phase space at =0 and the band of closed orbits disappears. reion where the closed orbits eist. At α 0 (β (d) For 0! β! we must consider the fied points at φ - / 0 arccos β * 0. So these points are saddles. Here is phase portrait: τ 0 (8) 8 sin φ () = (sqrt()/4) ( ) sin(phi) phi = (/) (beta cos(phi)/sqrt() cos(phi)/(8 sqrt())) beta = phi

7 3. The only fied point occurs at the oriin. Linearization tells us to epect a non-isolated fied point, but by plottin the nullclines we can see that this is not the case. From the plot of the vector field below, we can see that the inde around the fied point at the oriin is zero. = y y = + y y

8 (a) The only fied point is at the oriin. We d like to construct a Liapunov function, V(), such that V()>0 and dv dt <0 for all,. Consider V= ay with a>0. The first condition is automatically satisfied. Let s investiate the second: V ẋ ayẏ (0) y 3 ' ay y 3 () a y 4 ay 4 () So the second condition is satisfied if we choose a=. Our Liapunov function is V= y. A system that has a Liapunov function cannot have a closed orbit. Assume such a system did have a closed orbit and consider the point 0 on it. After one circuit the system would have to return to 0 and we would have V =0 (since V depends only on T position). But we know that V must also be iven by V 43 0 V dt where T is the period of the orbit. Since the closed orbit cannot intersect with the oriin, this tells us that V is strictly less than zero - a contradiction! So there can be no limit cycle for this system. (b) If we have a radient system, then f=- V we know: f y V y Now let s do a proof in the opposite direction. Start with f and y: 5 (3) V and =- y, so as lon as V is well-behaved V y 0 (4) y and interate over y f y dy6 d67 5 f d6 dy6 d6 (5) y dy6 (6) So if we define V y 8 3 f d6 3 y dy we are uaranteed that f=- V i.e. the system is a radient system. f f y =+ and =+ so y, the system is a radient system. and =- V y, Let s assume we have a limit cycle. If we take one trip around it and return to our startin point, we must have V=0 for the trip (since V must be sinle-valued). However, if T the cycle has period T, we can also write: V 3 0 Vdt T T "3 0 V : ẋ dt 3 0 ẋ dt. For a cycle to eist, we cannot have ẋ 0 (or else motion would stop) so we have shown that V must be strictly less than zero. This is a contradiction, so radient systems cannot have limit cycles. Since this system is a radient system, it can have no limit cycle. 8